39.4k views
1 vote
In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 55 acres. Suppose we randomly survey 44 farmers from 1940 The middle 50% of the distribution for X bar , the bounds of which form the distance represented by the IQR, lies between what two values?

1 Answer

2 votes

Final answer:

To find the values representing the middle 50% of the distribution for X bar, we need to calculate the first quartile (Q1) and third quartile (Q3) using the z-score formula and the properties of a standard normal distribution. The values of X can then be found by multiplying the z-scores by the standard deviation and adding the mean.

Step-by-step explanation:

The middle 50% of the distribution for X bar lies between two values, which can be determined by finding the first quartile (Q1) and third quartile (Q3). The IQR (interquartile range) is then calculated by subtracting Q1 from Q3. In this case, we know that X bar has a mean of 174 acres and a standard deviation of 55 acres. Since the distribution is assumed to be normal, we can use the properties of a standard normal distribution to find the values for Q1 and Q3. We can use the z-score formula: z = (x - mean) / standard deviation.

With the given information, we can calculate the z-scores for Q1 and Q3 and use the z-score table to find the corresponding percentiles. The middle 50% of the distribution for X bar is the area between these two percentiles. To find the values of X that correspond to these percentiles, we can use the formula: x = z * standard deviation + mean.

User Vinga
by
8.0k points